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Consider the next second order differential equations system:

$$x''=0,$$ $$y''=-\frac{1}{y} (x')^2,$$ with initial conditions $y(0)=y_0$, $x(0)=x_0$, $y'(0)=0$, $x'(0)=1$. I need to know how to solve this. Which method can I use? I really don't know how to start.

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    $\begingroup$ $x''=0\implies x'=c$ and sub into the second one and integrate. I think you can finish it on your own. $\endgroup$
    – cr001
    Nov 13 '15 at 5:42
  • $\begingroup$ @cr001 I can't. $\endgroup$
    – user29232
    Nov 13 '15 at 5:48
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First, integrate $x''$: $$ x' = \int x'' dt = \int 0 dt = C = 1 $$ Integrate again: $$ x = \int x' dt = \int 1 dt = t + C $$ Since $x(0) = x_0, C = x_0$. Therefore $x = t + x_0$.

Since $x' = 1$, $$ y'' = -\frac{1}{y}$$

There are no closed form solutions to $y$.

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  • $\begingroup$ Does that mean that the system doesn't have a solution? $\endgroup$
    – user29232
    Nov 13 '15 at 5:58
  • $\begingroup$ @Sajindia According to wolframalpha, it does. $\endgroup$ Nov 13 '15 at 5:59
  • $\begingroup$ But what is a closed form solution? $\endgroup$
    – user29232
    Nov 13 '15 at 6:01
  • $\begingroup$ @Sajindia Closed form solution $\endgroup$ Nov 13 '15 at 6:13

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